HDU 4738 Caocao's Bridges（双联通分量+并查集）

大意：有n座岛和m条桥，每条桥上有w个兵守着，现在要派不少于守桥的士兵数的人去炸桥，只能炸一条桥，使得这n座岛不连通，求最少要派多少人去。

思路：我们就是要缩点后直接求桥上人的最少数量。（PS：1、注意图如果不联通直接输出0。2、如果图中的桥上人为0，个那么要让一个人去。3、重边的问题。这里可以忽略）

#include<map>
#include<queue>
#include<cmath>
#include<cstdio>
#include<stack>
#include<iostream>
#include<cstring>
#include<algorithm>
#define LL int
#define inf 0x3f3f3f3f
#define eps 1e-8
#include<vector>
#define ls l,mid,rt<<1
#define rs mid+1,r,rt<<1|1

using namespace std;

const int Ma = 1100;
struct node{
    int to,w,next;
}q[Ma*Ma];

int head[Ma*Ma],dfn[Ma],num[Ma],du[Ma],stk[Ma*5],vis[Ma],low[Ma];
int cnt,top,tim,scc,out[Ma],f[Ma],n,mi;

void Add(int a,int b,int c){
    q[cnt].to = b;
    q[cnt].w = c;
    q[cnt].next = head[a];
    head[a] = cnt++;
}

void init(){
    scc =  cnt = top = 0;
    tim =  1;
    mi = inf;
    memset(head,-1,sizeof(head));
    for(int i = 1;i <= n;++ i){
        f[i] = i;
        low[i] = vis[i] = out[i] = num[i] = dfn[i] = 0;
    }
}

void Tarjan(int u,int To){
    low[u] = dfn[u] = tim++;
    vis[u] = 1;
    stk[top++] = u;
    for(int i = head[u]; ~i ; i = q[i].next){
        int v = q[i].to;
        if(i == (To^1)) continue;
        if(!vis[v]){
            Tarjan(v,i);
            low[u] = min(low[u],low[v]);
            if(low[v] > dfn[u])
                if(q[i].w < mi)
                    mi = q[i].w;
        }
        else
            low[u] = min(low[u],dfn[v]);
    }
    if(low[u] == dfn[u]){
        scc++;
        while(top > 0&&stk[top] != u){
            top --;
            vis[stk[top] ] = 1;
            num[stk[top] ] = scc;
        }
    }
}

int fi(int x){
    return f[x] == x ? x:f[x]=fi(f[x]);
}

void mer(int a,int b){
    int x = fi(a);
    int y = fi(b);
    x > y ? f[x] = y:f[y] = x;
}

int main(){
    int m,i,j,k,a,b,c,cla;
    while(~scanf("%d%d",&n,&m)){
        if(!n&&!m) break;
        init();
        for(i = 0;i < m;++ i){
            scanf("%d%d%d",&a,&b,&c);
                Add(a,b,c);
                Add(b,a,c);
                mer(a,b);
        }
        int tmp = 0;
        for(i = 1;i <= n;++ i)
            if(f[i] == i){
                tmp++;
                if(tmp > 1) break;
            }
        if(tmp > 1){
            puts("0");continue;
        }
        Tarjan(1,-1);
        for(i = 1;i <= n;++ i){
            for(j = head[i]; ~j ; j=q[j].next){
                int v = q[j].to;
                if(num[i]!=num[v]){
                    out[num[i]]++;
                }
            }
        }
        int ans = 0;
        for(i = 1;i <= scc;++ i){
            if(out[i]==0)
                ans++;
        }
        if(ans==1){
            puts("-1");
        }
        else if(!mi){
            puts("1");
        }
        else
            printf("%d\n",mi);
    }
    return 0;
}

HDU 4738 Caocao's Bridges（双联通分量+并查集）